THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
Jeffrey D. Vaaler*
José Felipe Voloch**
1. Introduction
Let k be a Galois extension of Q with [k : Q] = d ≥ 2. The purpose of this paper is
to give an upper bound for the least prime which does not split completely in k in terms
of the degree d and the discriminant ∆k . Our estimate improves on the bound given by
Lagarias, Montgomery and Odlyzko [3]. We note, however, that with the assumption of
the generalized Riemann hypothesis much stronger bounds have been obtained by Murty
[7]. In fact the analytic method employed in [7] can be used to produce an unconditional
bound of the same general type as ours. The case of an abelian extension was considered
earlier by Bach and Sorensen [1] and Oesterlé [8].
Our method is essentially elementary. It is based on an application of the product
α
formula to the binomial coefficient N
, where α is an irrational algebraic integer in k
and Tracek/Q (α) = 0. A similar idea has been used in [11] to give a lower bound on the
number of primes that do split completely in k. At one point in our argument we appeal to
the prime number theorem with an error term in which all constants are given explicitly.
Thus for each d we obtain a bound on the least prime which does not split completely
√
provided |∆k | is large compared with d. In the special case k = Q( p) a somewhat
* Research of the first author was supported in part by the National Science Foundation (DMS-9622556)
and the Texas Advanced Research Project,
** Research of the second author was supported in part by the National Security Agency (MDA90497-1-0037).
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1
2
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
simpler argument can be used which avoids the prime number theorem and leads to a
result valid for all discriminants. The simpler argument differs insignificantly from that
used by Gauss in the course of his first proof of quadratic reciprocity [2, art. 129].
THEOREM 1. If
exp(max{105, 25(log d)2 }) ≤ 8|∆k |1/2(d−1) ,
(1.1)
then there exists a prime p such that p does not split completely in k and
p ≤ 26d2 |∆k |1/2(d−1) .
(1.2)
2. Nonarchimedean estimates
Throughout this section we assume that all primes p ≤ dN (d − 1)−1 split in k, where
N ≥ d is a positive integer parameter. We further assume that α is a nonzero algebraic
integer in k with [Q(α) : Q] = δ and Tracek/Q (α) = 0. Then we write α = α1 , α2 , . . . , αδ
for the distinct conjugates of α in k and
f (x) =
δ
Y
(x − αi )
i=1
for the minimal polynomial of α over Q. Obviously f is a monic, irreducible polynomial
in Z[x] with 2 ≤ δ = deg(f ) and δ | d. We also define
AN (α) = Normk/Q
where
x
N
n α o
N
dN
= (−1)
−d
(N !)
−1
nNY
od/δ
f (n)
,
n=0
is the binomial coefficient. Clearly AN (α) is a nonzero rational number.
LEMMA 2. For each prime p with p ≤ dN (d − 1)−1 , the congruence
f (x) ≡ 0
mod p
has at least one root in the set {0, 1, 2, . . . , [(δ − 1)p/δ]}.
(2.1)
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
3
Proof. Let p be a prime with p ≤ dN (d − 1)−1 . All embeddings of k in an algebraic closure
Qp are contained in Qp . Hence all roots of f occur in Zp and therefore f splits in Z/pZ[x].
Let a1 , a2 , . . . , aδ be elements of {0, 1, 2, . . . , p − 1} such that
f (x) ≡
δ
Y
(x − ai )
mod p.
i=1
If p ≤ δ then the result is trivial, so we may assume that δ < p. Because Tracek/Q (α) = 0
we have
δ
X
ai ≡ 0
mod p.
(2.2)
i=1
Now assume, contrary to the statement of the lemma, that (δ − 1)p/δ < ai ≤ p − 1 for
each i = 1, 2, . . . , δ. Then we get
(δ − 1)p <
δ
X
ai ≤ δ(p − 1),
i=1
which contradicts (2.2)
LEMMA 3. The number AN (α) is a nonzero integer.
Proof. Let p be a prime with p ≤ N . As before all roots of f occur in Zp . It follows that
αi
N belongs to Zp for each i = 1, 2, . . . , δ. In particular the p-adic absolute value of AN (α)
satisfies |AN (α)|p ≤ 1 for each p ≤ N , and of course for p > N the bound |AN (α)|p ≤ 1 is
trivial. Thus AN (α) is a nonzero integer.
LEMMA 4. For each prime p such that N < p ≤ dN (d − 1)−1 , we have
log |AN (α)|p ≤ dδ −1 (− log p).
(2.3)
4
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
Proof. As log |N !|p = 0 for N < p, we find that
log |AN (α)|p = dδ −1
−1
n NX
o
log |f (n)|p − d log |N !|p
n=0
= dδ
≤ dδ
−1
−1
N
−1
X
∞
X
n=0
m=1
pm |f (n)
(− log p)
N
−1
X
(− log p)
(2.4)
1.
n=0
p|f (n)
By hypothesis we have d ≤ N < p ≤ dN (d − 1)−1 , and therefore
[(δ − 1)p/δ] ≤ [(d − 1)p/d] < (d − 1)p/d ≤ N.
By Lemma 2 the sum on the right of (2.4) contains at least one nonzero term and this
verifies (2.3).
THEOREM 5. If exp(max{105, 25(log d)2 }) ≤ N then
X
log |AN (α)|p ≤ −N δ −1 .
(2.5)
p
Proof. We use the explicit error term in the prime number theorem obtained by Rosser
and Schoenfeld [10, Theorem 2]. An easy consequence of their result is that
|
X
p
log p − X| ≤ (1/2)X exp{−(2/5) log X}
p≤X
for all X ≥ exp(105). It follows that
|
X
p≤X
log p − X| ≤
X
d(2d − 1)
(2.6)
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
5
whenever exp(max{105, 25(log d)2 }) ≤ X. Now let exp(max{105, 25(log d)2 }) ≤ N , set
X = dN (d − 1)−1 and = (d(2d − 1))−1 . Then (2.3) and (2.6) imply that
X
X
log |AN (α)|p ≤ dδ −1
(− log p)
p
N <p≤X
= dδ
−1
≤ dδ
−1
(N − X) + (
X
log p − N ) + (X −
p≤N
X
log p)
p≤X
{N − X + (N + X)}
= −N δ −1 .
3. Archimedean estimates
In this section we assume that f is a polynomial in R[x] with deg(f ) = M ≥ 1. Then
we write α1 , α2 , . . . , αL for the distinct roots of f in C so that
L
Y
f (x) = c0 (x − αi )e(l)
with e(l) ∈ {1, 2, . . . } and c0 6= 0.
l=1
It follows that the Mahler measure µ(f ) is given by
log µ(f ) = log |c0 | +
L
X
e(l) log+ |αl |.
l=1
We also require the norm
kf k∞ = sup{|f (z)| : z ∈ C, |z| ≤ 1}.
And we will use two well known inequalities (see [5, equation (4)], [6], or [9, Lemma 2])
2−M kf k∞ ≤ µ(f ) ≤ kf k∞
and
log µ(f 0 ) ≤ log M + log µ(f ).
By the square free kernel of f we understand the polynomial
L
Y
q(x) =
(x − αl ).
l=1
(3.1)
6
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
LEMMA 6. Let f be a polynomial in R[x], q the square free kernel of f , and
Bf = {β ∈ R : f 0 (β) = 0 and f (β) 6= 0}.
Then we have
|Bf | ≤ L − 1
X
and
log+ |β| ≤ log kqk∞ .
(3.2)
β∈Bf
Proof. There exists a polynomial p(x) in R[x] uniquely determined by the identity
L
p(x)
f 0 (x) X e(l)
=
=
.
f (x)
x − αl
q(x)
l=1
Clearly we have
Bf = {β ∈ R : p(β) = 0}.
¿From the identity f 0 (x)q(x) = f (x)p(x) we find that deg(p) = L − 1 and the leading
coefficient of p is M . Therefore |Bf | ≤ L − 1 and
log M +
X
log+ |β| ≤ log µ(p).
(3.3)
β∈Bf
Then the basic inequalities (3.1) imply that
log µ(p) + log µ(f ) = log µ(q) + log µ(f 0 )
(3.4)
≤ log kqk∞ + log M + log µ(f ).
The remaining inequality in (3.2) follows from (3.3) and (3.4).
LEMMA 7. Let f be a polynomial in R[x] and q the square free kernel of f . Then we
have
Z V
U
d
log+ |f (x)| dx ≤ M log+ |U | + log+ |V | + 2 log+ kqk∞ + 2L log+ kf k∞ . (3.5)
dx
Proof. Write Bf (U, V ) for the set of distinct roots of f 0 which occur in the interval (U, V )
and which are not roots of f . To begin with we will establish the inequality
Z V
X
d
log+ |f (x)| dx ≤ log+ |f (U )| + log+ |f (V )| + 2
log+ |f (β)|.
dx
U
β∈Bf (U,V )
(3.6)
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
7
Suppose that (u, v) ⊆ R is a bounded open interval such that
1 < f (x)
whenever
u < x < v.
(3.7)
Then let
Bf (u, v) = {β1 , β2 , . . . βJ },
and write
u = β0 < β1 < β2 < · · · < βJ < βJ+1 = v.
Obviously J = 0 in case Bf (u, v) is empty. In view of (3.7) we have
Z
J
X
d log+ |f (x)| dx =
dx
j=0
v
u
βj+1 Z
βj
J Z
X
=
j=0
0
f (x) dx
f (x)
βj+1
βj
f 0 (x)
dx
f (x)
J
X
log f (βj+1 ) − log f (βj )
=
(3.8)
j=0
≤
J
X
max log+ |f (βj+1 )|, log+ |f (βj )|
j=0
X
≤ log+ |f (u)| + log+ |f (v)| + 2
log+ |f (β)|.
β∈Bf (u,v)
It is clear that (3.8) continues to hold if
f (x) < −1
whenever
u < x < v.
Next we write
{x ∈ R : U < x < V, 1 < |f (x)|} =
K
[
(3.9)
(uk , vk ),
k=1
where {(uk , vk ) : k = 1, 2, . . . K} is a finite, disjoint collection of open intervals. Then we
have
Z
V
U
K
X
d
log+ |f (x)| dx =
dx
k=1
Z
vk uk
d log+ |f (x)| dx
dx
(3.10)
8
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
and so we can apply the estimate in (3.8) to each term in the sum on the right of (3.10).
In doing so we note that if U < uk < V then, since x → log+ |f (x)| is continuous, we have
log+ |f (uk )| = 0, and similarly if U < vk < V . It follows then that
Z
V
U
K
X
d
log+ |f (x)| dx =
dx
Z
k=1
vk uk
+
d log+ |f (x)| dx
dx
+
≤ log |f (U )| + log |f (V )| + 2
K
X
X
log+ |f (β)|
k=1 β∈B(uk ,vk )
X
= log+ |f (U )| + log+ |f (V )| + 2
log+ |f (β)|,
β∈Bf (U,V )
and this verifies the inequality (3.6).
In order to further estimate the terms on the right of (3.6) we employ the inequality
|f (z)| ≤ kf k∞ max(1, |z|)M ,
which follows easily from the maximum modulus theorem (see [9, Lemma 4]). Also, we
have
log+ |w1 w2 | ≤ log+ |w1 | + log+ |w2 |
for all pairs of complex numbers w1 and w2 . Combining these observations we find that
log+ |f (z)| ≤ log+ kf k∞ + M log+ |z|.
(3.11)
Now we can combine (3.2), (3.6), (3.11) and so establish the bound:
Z
V
U
d
log+ |f (x)| dx
dx
≤ M log+ |U | + M log+ |V | + (2|Bf | + 2) log+ kf k∞ + 2M
X
β∈Bf
≤ M log+ |U | + M log+ |V | + 2L log+ kf k∞ + 2M log+ kqk∞
This proves the lemma.
log+ |β|.
(3.12)
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
9
LEMMA 8. Let f be a polynomial in Z[x], q the square free kernel of f , deg(f ) = M
and deg(q) = L. Let N be a positive integer and assume that f has no roots in the set
{0, 1, 2, . . . , N − 1}. Then we have
−1
NX
log |f (n)|−
N
Z
log |f (x)| dx
(3.13)
0
n=0
≤ M (2 + log N ) + (L + 12 ) log kf k∞ + M log kqk∞ .
Proof. From a standard application of Stieltjes integration we obtain the identity
N
−1
X
log |f (n)| =
n=0
N
−1
X
log+ |f (n)|
n=0
N−
Z
log+ |f (x)| d{[x] + 12 }
=
0−
N
Z
log+ |f (x)| dx −
=
0
N
Z
+
B1 (x)
0
1
2
(3.14)
log+ |f (N )| +
1
2
log+ |f (0)|
d
log+ |f (x)| dx.
dx
Here
B1 (x) = x − [x] −
1
2
when x ∈
/ Z, and B1 (x) = 0 when x ∈ Z,
is the first periodic Bernoulli polynomial. We use the bound |B1 (x)| ≤
1
2,
the estimate
(3.5) from the previous lemma, (3.11) and (3.14). In this way we arrive at the inequality
−1
NX
log |f (n)| −
N
Z
log+ |f (x)| dx
0
n=0
≤
1
2
+
+
max log |f (0)|, log |f (N )| +
≤ M log N + (L +
+
1
2 ) log
1
2
Z
0
+
N
d log+ |f (x)| dx
dx
kf k∞ + M log kqk∞ .
(3.15)
10
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
Next we observe that
Z
Z
N
log |f (x)| dx−
0
N
log+ |f (x)| dx
0
Z
N
log− |f (x)| dx
=
≤
0
L
X
Z
e(l)
R
l=1
≤
L
X
Z
e(l)
=
(3.16)
log− |x − <(αl )| dx
R
l=1
L
X
log− |x − αl | dx
Z
e(l)
l=1
log− |x| dx
R
= 2M.
Then we combine (3.15) and (3.16). We find that
−1
NX
log |f (n)|−
n=0
Z
N
log |f (x)| dx
(3.17)
0
+
+
≤ M (2 + log N ) + (L + 12 ) log kf k∞ + M log kqk∞ .
To complete the proof we note that 1 ≤ kf k∞ and 1 ≤ kqk∞ , because both f and q belong
to Z[x].
We define ρ : C → R by
ρ(z) =
1
2
Z
1
log |t − z| dt + 1.
(3.18)
−1
If follows from the basic theory of logarithmic potential functions (see [3, Appendix 4, §1.])
that ρ is nonnegative, continuous, subharmonic, and the restriction of ρ to C \ [−1, 1] is
harmonic. For our purposes we require information about ρ in the closed unit disc.
LEMMA 9. For all complex z = x + iy with |z| ≤ 1 we have
ρ(z) ≤
π|y|
+ (log 2)|z|2 .
2
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
11
Proof. Let ψ(z) be defined in the closed unit disc by
ψ(z) =
∞
X
z 2n
.
2n(2n
−
1)
n=1
In the upper half-disc {z ∈ C : 0 < =(z), |z| < 1} we find that
1
2
Z
1
−1
log(z − t) dt + 1 = 12 (1 − z) log(z − 1) + 12 (1 + z) log(z + 1)
∞
πi πiz X
z 2n
=
−
+
2
2
2n(2n − 1)
n=1
=
(3.19)
πi πiz
−
+ ψ(z),
2
2
where we have used the principal branch of the logarithm. In the lower half-disc {z ∈ C :
0 > =(z), |z| < 1} the corresponding identity is
1
2
Z
1
log(z − t) dt + 1 =
−1
−πi πiz
+
+ ψ(z).
2
2
(3.20)
Then (3.19), (3.20) and the continuity of ρ imply that
ρ(z) = <
n Z
1
2
1
o π|y|
+ < ψ(z) ,
log(z − t) dt + 1 =
2
−1
(3.21)
at all points z in the closed unit disc. By the maximum modulus theorem
ψ(z) |ψ(z)| = |z|2 2 ≤ |z|2 kψk∞ = (log 2)|z|2
z
(3.22)
for all z in the closed unit disc. The lemma plainly follows from (3.21) and (3.22).
4. The existence of special numbers
Here we assume that k is an algebraic number field having degree d ≥ 2 over Q. Let
σ1 , σ2 , . . . , σd be the distinct embeddings of k into C. We assume that σ1 , σ2 , . . . , σr are
real, that σr+1 , σr+2 , . . . , σr+s are complex and not real, and that σ r+j = σr+s+j for
j = 1, 2, . . . , s. We write Ok for the ring of integers in k and ∆k for the discriminant.
12
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
THEOREM 10. There exists a nonzero algebraic integer α in k such that
Tracek/Q (α) = 0
max |σi (α)| ≤ 4|∆k |1/2(d−1) .
and
1≤i≤d
(4.1)
Moreover, if [Q(α) : Q] = δ and f in Z[x] is the minimal polynomial of α over Q, then
kf k∞ ≤ 8δ |∆k |δ/2(d−1) .
(4.2)
Proof. To begin with we observe that the set of algebraic integers in Ok which satisfy
max |σi (α)| ≤ T
1≤i≤d
is finite for every positive T . Thus it suffices to prove that for every > 0 there exists an
algebraic integer α in k such that
Tracek/Q (α) = 0
and
max |σi (α)| ≤ (4 + )|∆k |1/2(d−1) .
1≤i≤d
Let ω1 , ω2 , . . . , ωd be an integral basis for Ok . We write Ω for the d × d matrix Ω =
(σi (ωj )), where i = 1, 2, . . . , d indexes rows and j = 1, 2, . . . , d indexes columns. Then we
define W to be the d × d matrix which is organized into blocks as

1r
W = 0
0
0
1
2 1s
1
2i 1s

0
1
,
2 1s
−1
2i 1s
where 1r and 1s are identity matrices. We note that (det Ω)2 = ∆k , det W = (−2i)−s and
the product W Ω is a d × d matrix with real entries. Next we define
d
Λk = λ ∈ Z :
d
X
Tracek/Q (ωj )λj ≡ 0
mod d .
j=1
As Λk is the kernel of the group homomorphism λ →
Pd
j=1
Tracek/Q (ωj )λj from Zd into
Z/dZ, it follows that Λk ⊆ Zd is a sublattice of index at most d.
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
13
We now assume that 1 ≤ r and let t denote a positive parameter. Then we define
n
Cr,s (t) = y ∈ Rd : |y1 | < 1, |yi | ≤ t if 2 ≤ i ≤ r,
o
(4.3)
and (yr+j )2 + (yr+s+j )2 ≤ t2 if 1 ≤ j ≤ s .
It is clear that Cr,s (t) is a convex, symmetric subset of Rd , and a simple computation
shows that
Vold {Cr,s (t)} = 2d (π/4)s td−1 .
(4.4)
Hence the linear image
Kr,s (t) = (W Ω)−1 Cr,s (t) = x ∈ Rd : W Ωx ∈ Cr,s (t) ,
is also a convex, symmetric subset. And using (4.4) we find that
Vold Kr,s (t) = Vold (W Ω)−1 Cr,s (t)
= | det W Ω|−1 Vold {Cr,s (t)} = 2d (π/2)s td−1 |∆k |−1/2 .
(4.5)
Let 0 < η and set t = (2 + η)|∆k |1/2(d−1) . Then
Vold Kr,s (t) = 2d (2 + η)d−1 (π/2)s > [Zd : Λk ]2d ,
and so by Minkowski’s convex body theorem there exists a nonzero point ξ in Kr,s (t) ∩ Λk .
Pd
Using ξ we define β = j=1 ξj ωj , so that β is a nonzero point in Ok . ¿From the definitions
of W , Ω and Cr,s (t) we find that
|σ1 (β)| < 1 and |σi (β)| ≤ (2 + η)|∆k |1/2(d−1) for i = 2, 3, . . . , d.
(4.6)
It is clear from the first inequality on the left of (4.6) that β ∈ Ok \ Z. ¿From the definition
of Λk we learn that
Tracek/Q (β) = md
with m ∈ Z.
We conclude that α = β − m also belongs to Ok \ Z and Tracek/Q (α) = 0. We also get the
estimate
d
−1 X
|m| = d
σi (β) ≤ max |σi (β)| ≤ (2 + η)|∆k |1/2(d−1)
i=1
1≤i≤d
14
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
and therefore
max |σi (α)| ≤ (4 + 2η)|∆k |1/2(d−1) .
1≤i≤d
In view of our previous remarks, this verifies the inequality on the right of (4.1).
Next we assume that r = 0 and define
n
C0,s (t) = y ∈ Rd : (y1 )2 + t−2 (ys+1 )2 < 1,
2
2
and (yj ) + (ys+j ) ≤ t
2
o
if 2 ≤ j ≤ s .
(4.7)
Clearly C0,s (t) is also a convex, symmetric subset of Rd , and again we have
Vold {C0,s (t)} = 2d (π/4)s td−1 .
(4.8)
We set t = (2 + η)|∆k |1/2(d−1) and proceed as before to determine a nonzero point β in
Ok . In this case we find that
(<(σ1 (β))2 + t−2 (=(σ1 (β))2 < 1,
1/2(d−1)
and |σj (β)| ≤ (2 + η)|∆k |
(4.9)
for 2 ≤ j ≤ s.
The first inequality on the left of (4.9) shows that β ∈ Ok \ Z. The rest of the argument
verifying (4.1) is essentially the same.
To complete the proof, let f in Z[x] be the minimal polynomial of α over Q. Then (4.1)
implies that the Mahler measure µ(f ) satisfies the bound
µ(f ) ≤ 4δ |∆k |δ/2(d−1) ,
where
[Q(α) : Q] = δ.
And from the inequality on the left of (3.1) we conclude that
kf k∞ ≤ 2δ µ(f ) ≤ 8δ |∆k |δ/2(d−1) .
THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
15
5 Proof of Theorem 1
Let k be a Galois extension of Q with [k : Q] = d ≥ 2. As in section 2 we assume that
all primes p ≤ dN (d − 1)−1 split in k, where N ≥ d is a positive integer parameter. By
Theorem 10 there exists an algebraic integer α in Ok \ Z with Tracek/Q (α) = 0 and
max |αi | ≤ 4|∆k |1/2(d−1) ,
(5.1)
1≤i≤δ
where [Q(α) : Q] = δ and α = α1 , α2 , . . . , αδ are the conjugates of α in k. We assume that
exp(max{105, 25(log d)2 }) ≤ 8|∆k |1/2(d−1) ≤ N.
(5.2)
Then the minimal polynomial f of α over Q satisfies the bound (4.2) and therefore kf k∞ ≤
N δ . Let AN (α) be defined as in (2.1). Then Theorem 5 and the product formula imply
that
0 = log |AN (α)| +
X
log |AN (α)|p ≤ log |AN (α)| − N δ −1 .
(5.3)
p
And from Lemma 8 we get the estimate
log |AN (α)| = dδ −1
N
−1
X
log |f (n)| − d log N !
n=0
≤ dδ −1
nZ
N
log |f (x)| dx + δ(2 + log N ) + (2δ + 12 ) log kf k∞
0
o
(5.4)
− d{N log N − N }
≤ dN δ
−1
nZ
1
log |N
−δ
o
f (N x)| dx + δ + 6d2 log N
0
Combining (5.3) and (5.4) we obtain the inequality
nZ 1
o 6d3 log N .
1≤d
log |N −δ f (N x)| dx + δ +
N
0
(5.5)
Next we derive (5.5) again but with −α in place of α, and then we combine the two bounds.
In this way we establish the estimate
n Z 1
o 6d3 log N −δ
1
1≤d 2
log |N f (N x)| dx + δ +
.
N
−1
(5.6)
16
JEFFREY D. VAALER, JOSÉ FELIPE VOLOCH
¿From Lemma 9 and (5.1) we get
n Z
1
2
1
log |N
−δ
o
f (N x)| dx + δ =
δ n Z
X
1
2
−1
i=1
=
δ
X
1
o
log |t − N −1 αi | dt + 1
−1
ρ(N −1 αi )
i=1
≤ 3N
−1
≤ 12δN
δ
X
(5.7)
|αi |
i=1
−1
|∆k |1/2(d−1) .
Thus (5.2), (5.6) and (5.7) lead to the bound
12d2 |∆ |1/2(d−1) 6d3 log N k
+
N
N
12d2 |∆ |1/2(d−1) 6d3 max{105, 25(log d)2 } k
≤
+
N
exp(max{105, 25(log d)2 })
12d2 |∆ |1/2(d−1) k
≤
+ 10−40 .
N
1≤
(5.8)
We select
N = 13d2 |∆k |1/2(d−1) ,
use the hypothesis (5.2), and obtain a contradiction to (5.8). We have shown that if
exp(max{105, 25(log d)2 }) ≤ 8|∆k |1/2(d−1) ,
(5.9)
then there exists a prime number p with
p ≤ dN (d − 1)−1 ≤ 26d2 |∆k |1/2(d−1)
such that p does not split completely in k.
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THE LEAST NONSPLIT PRIME IN GALOIS EXTENSIONS OF Q
17
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Jeffrey D. Vaaler, Department of Mathematics, University of Texas, Austin, TX 78712,
e-mail: [email protected]
José Felipe Voloch, Department of Mathematics, University of Texas, Austin, TX 78712,
e-mail: [email protected]